If P(x, y),F_1 = (3, 0),F_2 = (-3, 0) and 16x | vedclass

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(C) The given equation of the curve is $16x^2 + 25y^2 = 400$.Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$,which is the standard equa

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$10$

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(C) The given equation of the curve is $16x^2 + 25y^2 = 400$.Dividing by $400$,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$,which is the standard equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 = 25$ and $b^2 = 16$.Here,$a = 5$ and $b = 4$.The foci are given by $(\pm ae, 0)$.We calculate eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.Thus,the foci are $(\pm 5 	imes \frac{3}{5}, 0) = (\pm 3, 0)$,which matches $F_1 = (3, 0)$ and $F_2 = (-3, 0)$.By the definition of an ellipse,the sum of the distances of any point $P$ on the ellipse from the two foci is equal to the length of the major axis,which is $2a$.Therefore,$PF_1 + PF_2 = 2a = 2 	imes 5 = 10$.

If $P(x, y)$,$F_1 = (3, 0)$,$F_2 = (-3, 0)$ and $16x^2 + 25y^2 = 400$,then $PF_1 + PF_2 = \dots$

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